A point mass m is moved in a vertical circle of radius r with the help of a string. The velocity of the mass is  $\sqrt{7\mathit{\text{gr}}}$ at the lowest point. The tension in the string at the lowest point is 

Velocity of point mass in vertical circle at lowest point, ${\text{V}}_1=\sqrt{7\text{gr}}$

$\therefore {\text{\hspace{0.17em}\hspace{0.17em}V}}_{\text{1}}=\sqrt{7\text{\hspace{0.17em}gr}}>\sqrt{5\text{\hspace{0.17em}gr}}$

Hence, point mass will have completed the vertical circular path
We know that,

${\text{T}}_{\text{bottom}}-\text{mg}=\frac{{\text{mυ}}^{\text{2}}}{\text{r}}=\frac{\text{m}}{\text{r}}\left(\sqrt{7\text{\hspace{0.17em}gr}}\right)$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}T}}_{\text{bottom}}-\text{mg}=7\text{\hspace{0.17em}mg}$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}T}}_{\text{bottom}}=8\text{\hspace{0.17em}mg}$